R0053-27374 · q-20261015-advance-20261118-r07-pumping-lemma-regular-languages-0049
exists p forall w exists split forall i量詞與split constraints責任。操作=pump-dfa-loop-witness；資料={"modulus":24,"word_length":96}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。
G001 — exists p forall w exists split forall i量詞與split constraints責任。操作=pump-dfa-loop-witness；資料={"modulus":24,"word_length":96}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。
操作前
source roles = [pump-dfa-loop-witness]
操作後
derived = [[modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified]]
條件：無
此頁直接列完整來源欄序與mode特定角色，不靠JavaScript或由答案外形猜介面；No、undefined、infinity各保原意。
規則：source_ordered_fields_without_JS · 依賴：G001
操作前
Pumping ORIGINAL unary z length := sub(96, 24)
操作後
Pumping ORIGINAL unary z length = 72
條件：無
逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。
規則：exact_rational_arithmetic · 依賴：G001, S001
操作前
source roles = [24, 96]
操作後
derived = [[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 17], [17, 18], [18, 19], [19, 20], [20, 21], [21, 22], [22, 23], [23, 0]], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 0], [0, 24, 0], 0, 24, 72]
條件：無
DFA狀態是長度mod m，逐每個狀態的unary transition加1取mod；前m+1狀態在位置0,m同為0。原word length真能被m整除，故x空、y長m、z為餘段給合法且接受的w。
規則：Pumping_complete_unary_DFA_cycle_and_actual_accepted_length · 依賴：G001, S002
操作前
source roles = [24, 96, 72]
操作後
derived = [EVERY integer i>=0: pumped length=i*modulus+z, z is divisible by modulus, state remains0 after each loop;accepted, 是]
條件：無
任意i>=0重走同一完整loop i次回0，再讀z；因z亦為m倍數，終態仍0。以完整transition與全稱不變量支持所有i，不是只測若干pump次數。
規則：Pumping_ALL_i_loop_invariant_NOT_finite_pump_tests · 依賴：G001, S003
S004 · premise_and_result_audit — 回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。S003 · premise_and_result_audit — 回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。S002 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL unary z length，不以小數近似替代。S001 · premise_and_result_audit — 回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。(24, 96, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 0], [0, 24, 0], 0, 24, 72, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, 是)
依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。
覆核日期：2026-10-26
原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。
input commit: 0f7be61e2c22317a4075561dc822c87d881a50dboriginal question SHA-256: b2e72265fe815ba60d1d13b6836fb8149dc96db1ac52def3556f20369cb9cf12candidate record SHA-256: aa30df68865ed41b03bd689f6efe154d451dc08722734f536d61aebc6de314c4semantic review receipt SHA-256: 1b7a62514347b82518305d5feb4369a52d7d44c442a0f2b131e9eb3cd7b45f62question-first witness SHA-256: 79dc2a1077081f3ea2a03f75e32778163bf05e814cdf4c29ed86e567295f9ee1Lean: NOT_RUN · human review: NOT_PERFORMED